Answer:
[tex]\boxed {\boxed {\sf 339 \ K }}[/tex]
Explanation:
The question asks us to find the new temperature given a change in pressure. We will use Gay-Lussac's Law, which states the pressure of a gas is directly proportional to the temperature. The formula is:
[tex]\frac {P_1}{T_1}=\frac{P_2}{T_2}[/tex]
The pressure changes from 767 millimeters of mercury (P₁) to 800 millimeters of mercury (P₂).
[tex]\frac {767 \ mm \ Hg }{ T_1}= \frac{800 \ mm \ Hg}{ T_2}[/tex]
The temperature is initially 325 K (T₁), but we don't know the final temperature or T₂.
[tex]\frac {767 \ mm \ Hg }{ 325 \ K}= \frac{800 \ mm \ Hg}{ T_2}[/tex]
We are solving for the final temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator by the second denominator, then the first denominator by the second numerator.
[tex]767 \ mm \ Hg * T_2 = 325 \ K * 800 \ mm \ Hg[/tex]
T₂ is being multiplied by 767 millimeters of mercury. The inverse of multiplication is division. Divide both sides by 767 mm Hg.
[tex]\frac {767 \ mm \ Hg * T_2}{767 \ mm \ Hg} = \frac {325 \ K * 800 \ mm \ Hg }{767 \ mm \ Hg}[/tex]
[tex]T_2 = \frac {325 \ K * 800 \ mm \ Hg }{767 \ mm \ Hg}[/tex]
The units of millimeters of mercury (mm Hg) cancel.
[tex]T_2= \frac {325 \ K * 800 }{767 }[/tex]
[tex]T_2= \frac {260,000 }{767} \ K[/tex]
[tex]T_2= 338.983050847 \ K[/tex]
The original measurements have 3 significant figures, so our answer must have the same. For the number we found, that is the ones place. The 9 in the tenths place tells us to round the 8 up to a 9.
[tex]T_2 \approx 339 \ K[/tex]
The temperature changes from 325 Kelvin to 339 Kelvin.
